Ahockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of 4.0 m/s, skates by with the puck. after 1.00 s, the first player makes up his mind to chase his opponent. if he accelerates uniformly at 0.12 m/s2, determine each of the following. (a) how long does it take him to catch his opponent? (assume the player with the puck remains in motion at constant speed.) s (b) how far has he traveled in that time? m

Time it takes for player 1 to catch player 2: t = 7,55s

The players traveled a distance of: d = 34.2m

Explanation:

Problem Analysis

When player 1 catches player 2 they will both have traveled the same distance d and player 2 will have spent 1 second more time, then:

d₁ = d₂ = d

t₂ = t₁ + 1

Player kinematics 1

Player 1 moves with uniformly accelerated movement:

(Formula 1)

d: displacement in meters (m)

vi: initial speed = 0

a: acceleration = 0.12m/s²

t₂ = time in s  

We replace data in formula (1)

d = (1/2)(0,12)×t₁²

d = 0,6t₁² equation (1)

Player kinematics 2

Player 2 moves with constant speed:

d = v × t₂ Formula (2)

d: displacement in meters (m)

v = speed in m/s = 4 m/s

t₂: time in s

t₂ = t₁+1

We replace data in formula (2)

d = 4 × (t₁+1)

d = 4t₁ + 4 equation (2)

equation (1) = equation (2)

0,6t₁² = 4t₁ + 4

0,6t₁² - 4t₁ - 4 = 0

Applying the quadratic formula:

a = 0,6

b = -4

c = -4

We take value of the positive time because the negative time does not exist

t₁=7,55s

We replace in the equation (1) to calculate the distance traveled

d = 0,6(7,55)² = 34.2m

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)